Download Unit 21* Capacitance in AC Circuits

Unit 21
Capacitance in AC Circuits
Unit 21 Capacitance in AC Circuits
Objectives:
• Explain why current appears to flow through
a capacitor in an AC circuit.
• Discuss capacitive reactance.
• Discuss the relationship of voltage and
current in a pure capacitive circuit.
Unit 21 Capacitance in AC Circuits
Capacitive Reactance
Countervoltage limits the flow of current.
Unit 21 Capacitance in AC Circuits
• Capacitors appear to pass alternating current (AC).
• This is due to the capacitor charging for half of the
cycle and discharging for half of the cycle.
• This repetitive charging and discharging allows current
to flow in the circuit.
Unit 21 Capacitance in AC Circuits
Capacitive Reactance
• Capacitive reactance opposes the flow of
electricity in a capacitor circuit.
• Capacitive reactance (XC) is measured in
ohms (Ω).
• XC = 1 /(2fC)
• f = frequency
• C = capacitance
Unit 21 Capacitance in AC Circuits
Phase Relationships
• Capacitive current leads the applied
voltage by 90°.
• Inductive current lags the applied voltage
by 90°.
• Resistive current is in phase with the
applied voltage by 90.
Unit 21 Capacitance in AC Circuits
Capacitive current leads the applied voltage by
90.
Unit 21 Capacitance in AC Circuits
Power Relationships
• Capacitive power is measured in VARsC.
• Capacitive VARsC and inductive VARsL are
180° out of phase with each other.
• Capacitive VARsC and inductive VARsL
cancel each other.
Unit 21 Capacitance in AC Circuits
Capacitive power phase relationships.
Unit 21 Capacitance in AC Circuits
Inductive VARs and Capacitive VARs.
Unit 21 Capacitance in AC Circuits
Frequency Relationships
• Capacitive reactance (XC) is inversely
proportional to frequency (f).
• As frequency increases ↑ then capacitive
reactance decreases ↓.
• As frequency decreases ↓ then capacitive
reactance increases ↑.
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
C1 = 10 µF
C2 = 30 µF
C3 = 15 µF
Solving a sample series capacitor circuit:
Three capacitors (10µF, 30 µF, and 15 µF) are series
connected to a 480-V, 60-Hz power source.
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
C1 = 10 µF
XC1 = 265 Ω
C2 = 30 µF
C3 = 15 µF
First, calculate each capacitance reactance.
Remember (XC) = 1/(2пFC) and 2пF = 377 at 60 Hz.
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
C1 = 10 µF
XC1 = 265 Ω
C2 = 30 µF
XC2 = 88 Ω
C3 = 15 µF
XC3 = 177 Ω
XC1 = 1/ (377 x 0.000010) = 265.25 Ω
XC2 = 1/ (377 x 0.000030) = 88.417 Ω
XC3 = 1/ (377 x 0.000015) = 176.83 Ω
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
C1 = 10 µF
XC1 = 265 Ω
C2 = 30 µF
XC2 = 88 Ω
C3 = 15 µF
XC3 = 177 Ω
Next, find the total capacitive reactance.
XCT = XC1 + XC2 + XC3
XCT = 265.25 + 88.417 + 176.83 = 530.497 Ω
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
IT = 0.905 A
C1 = 10 µF
XC1 = 265 Ω
EC1 = ?
C2 = 30 µF
XC2 = 88 Ω
EC2 = ?
C3 = 15 µF
XC3 = 177 Ω
EC3 = ?
Next, calculate the total current.
IT = ECT / XCT
IT = 480 V / 530.497 Ω = 0.905 A
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
IT = 0.905 A
C1 = 10 µF
XC1 = 265 Ω
I1 = .905 A
EC1 = ?
C2 = 30 µF
XC2 = 88 Ω
I2 = .905 A
EC2 = ?
C3 = 15 µF
XC3 = 177 Ω
I3 = .905 A
EC3 = ?
Next, calculate the component voltage drops.
The total current flows through each component.
EC = IT x XC
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
IT = 0.905 A
C1 = 10 µF
XC1 = 265 Ω
I1 = .905 A
EC1 = 240 V
C2 = 30 µF
XC2 = 88 Ω
I2 = .905 A
EC2 = 80 V
C3 = 15 µF
XC3 = 177 Ω
I3 = .905 A
EC3 = 160 V
EC1 = .905 x 265.25 = 240.051 V
EC2 = .905 x 88.417 = 80.017 V
EC3 = .905 x 176.83 = 160.031 V
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
IT = 0.905 A
434 VARsCT
C1 = 10 µF
XC1 = 265 Ω
I1 = .905 A
EC1 = 240 V
? VARsC1
C2 = 30 µF
XC2 = 88 Ω
I2 = .905 A
EC2 = 80 V
? VARsC2
C3 = 15 µF
XC3 = 177 Ω
I3 = .905 A
EC3 = 160 V
? VARsC3
Now the reactive power can easily be computed!
Use the Ohm’s law formula.
VARsC = EC x IC = 480 x 0.905 = 434.4
Unit 21 Capacitance in AC Circuits
ET = 480 V
F = 60 Hz
XCT = 530.5 Ω
IT = 0.905 A
434 VARs
C1 = 10 µF
XC1 = 265 Ω
I1 = .905 A
EC1 = 240 V
217 VARsC1
C2 = 30 µF
XC2 = 88 Ω
I2 = .905 A
EC2 = 80 V
72 VARsC2
C3 = 15 µF
XC3 = 177 Ω
I3 = .905 A
EC3 = 160 V
144 VARsC3
VARsC1 = 240.051 x .905 = 217.246
VARsC2 = 80.017 x .905 = 72.415
VARsC3 = 160.031 x .905 = 144.828
Unit 21 Capacitance in AC Circuits
ET = ? V
F = 60 Hz
XCT = ? Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = ? Ω
I1 = ? A
EC1 = ? V
? VARsC1
C2 = 75 µF
XC2 = ? Ω
I2 = ? A
EC2 = ? V
? VARsC2
C3 = 20 µF
XC3 = ? Ω
I3 = ? A
EC3 = ? V
? VARsC3
Three capacitors (50 µF, 75 µF, and 20 µF) are
connected to a 60-Hz line. The total reactive power
is 787.08 VARsC.
Unit 21 Capacitance in AC Circuits
ET = ? V
F = 60 Hz
XCT = ? Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = 53 Ω
I1 = ? A
EC1 = ? V
? VARsC1
C2 = 75 µF
XC2 = ? Ω
I2 = ? A
EC2 = ? V
? VARsC2
C3 = 20 µF
XC3 = ? Ω
I3 = ? A
EC3 = ? V
? VARsC3
First, calculate the capacitive reactance (XC).
XC1 = 1/ 2πFC = 1/ 377 x .000050 = 53.05 Ω
XC1 = 53.05 Ω
Unit 21 Capacitance in AC Circuits
ET = ? V
F = 60 Hz
XCT = ? Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = ? A
EC1 = ? V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = ? V
? VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = ? V
? VARsC3
XC2 = 1/ 2πFC = 1/ 377 x .000075 = 35.367 Ω
XC3 = 1/ 2πFC = 1/ 377 x .000020 = 132.63 Ω
Unit 21 Capacitance in AC Circuits
ET = ? V
F = 60 Hz
XCT = 18 Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = ? A
EC1 = ? V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = ? V
? VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = ? V
? VARsC3
1/ XCT = 1/ XC1 + 1/ XC2 + 1/ XC3
1/ XCT = 1/ 53.05 + 1/ 35.367 + 1/ 132.63
XCT = 18.295 Ω
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = ? A
EC1 = ? V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = ? V
? VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = ? V
? VARsC3
Now use the formula: ET = √(VARsCT x XCT).
ET = √(787.08 x 18.295)
ET = 120 V
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = ? A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = ? A
EC1 = 120 V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = 120 V
? VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = 120 V
? VARsC3
All the voltage drops equal the source voltage.
ET = EC1 = EC2 = EC3 = 120 V
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = ? A
EC1 = 120 V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = 120 V
? VARsC2
Next, find the current: IT = ECT / XCT.
IT = 120 / 18.295
IT = 6.559 A
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = 120 V
? VARsC3
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = ? A
EC2 = 120 V
? VARsC2
Similarly: I1 = EC1 / XC1
I1 = 120 / 53.05
I1 = 2.262 A
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = 120 V
? VARsC3
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
? VARsC2
Similarly: I2 = EC2 / XC2
I2 = 120 / 35.367
I2 = 3.393 A
C3 = 20 µF
XC3 = 132 Ω
I3 = ? A
EC3 = 120 V
? VARsC3
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
? VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
? VARsC2
Similarly: I3 = EC3 / XC3
I3 = 120 / 132.63
I3 = 0.905 A
C3 = 20 µF
XC3 = 132 Ω
I3 = 0.905 A
EC3 = 120 V
? VARsC3
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
271 VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
? VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = 0.905 A
EC3 = 120 V
? VARsC3
Reactive power for each component is computed.
VARsC1 = EC1 x IC1
VARsC1 = 120 x 2.262 = 271.442
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
271 VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
407 VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = 0.905 A
EC3 = 120 V
? VARsC3
Reactive power for each component is computed.
VARsC2 = EC2 x IC2
VARsC2 = 120 x 3.393 = 407.159
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
271 VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
407 VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = 0.905 A
EC3 = 120 V
109 VARsC3
Reactive power for each component is computed.
VARsC3 = EC3 x IC3
VARsC3 = 120 x 0.905 = 108.573
Unit 21 Capacitance in AC Circuits
ET = 120 V
F = 60 Hz
XCT = 18 Ω
IT = 6.559 A
787 VARsC
C1 = 50 µF
XC1 = 132 Ω
I1 = 2.262 A
EC1 = 120 V
271 VARsC1
C2 = 75 µF
XC2 = 35 Ω
I2 = 3.393 A
EC2 = 120 V
407 VARsC2
C3 = 20 µF
XC3 = 132 Ω
I3 = 0.905 A
EC3 = 120 V
109 VARsC3
A quick check is done by adding the individual VARs
and comparing the value to the original VARs.
VARsC = 271.442 + 407.129 + 108.573 = 787.174
Unit 21 Capacitance in AC Circuits
Review:
1. Current appears to flow through a capacitor
in an AC circuit.
2. A capacitor appears to allow current flow
because of the periodic rise and fall of
voltage and current.
3. Current flow in a pure capacitive circuit is
only limited by capacitive reactance.
Unit 21 Capacitance in AC Circuits
Review:
4. Capacitive reactance is proportional to
the capacitance and frequency.
5. Capacitive reactance is measured in
ohms.
6. Current flow in a pure capacitive circuit
leads the voltage by 90.
Unit 21 Capacitance in AC Circuits
Review:
7. In a pure capacitive circuit, there is no
true power (watts).
8. Capacitive power is reactive and is
measured in VARs, as is inductance.
9. Capacitive and inductive VARs are 180
out of phase with each other.